Calendar 4 - Free MCQ Practice Test with solutions, UPSC Prelims Paper

15^th August, 1947 = (1946 years + Period from 1^st Jan., 1947 to 15th )
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1) ] odd days
= 58 odd days
= 2 odd days
Jan   Feb   Mar   Apr   May   Jun   Jul   Aug
= 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days
= (32 weeks + 3 days)
= 3,
Total number of odd days
= (0 + 1 + 2 + 3) odd days
= 6 odd days.
Hence, the required day was 'Saturday'.

We will show that the number of odd days between last day of February and last day of October is zero.
March   April   May   June   July   Aug.   Sept.   Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days
= 35 weeks
= 0 odd day
Number of odd days during this period = 0.
Thus, 1^st March of an year will be the same day as 1^st November of that year. Hence, the result follows

350 days = (350/7 = 50 weeks) i.e No odd days,
So it will be a Saturday.

We already know that the calendar after a leap year repeats again after 28 years.
Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

We know that, After every 400 years, the same day occurs.
Thus, if 9^th August 2016 is Saturday, before 400 years
i.e., on 9^th August 1616 has to be Saturday.

Given that the month begins on a Friday and has 31 days
Sundays = 3^rd, 10^th, 17^th, 24^th, 31^st
⇒ Total Sundays = 5
Every second & fourth Saturday is holiday. 2^nd & 4^th Saturday in every month = 2
Total days in the month = 31
Total working days = 31 - (5 + 2) = 24 days.

The year 1996 is a leap year. So, it has 2 odd days. But 17 th March  comes after 29 th February. So, the day on 17 th March, 1997 will be 1 day  beyond the day on 17 th March, 1996.

Here 17 th March, 1997 is Monday. So, 17 th March, 1996 is a Sunday.

When a century year leaves a remainder 0, when divided by 400 then it is a leap year (366 days).
So, 1200 has 366 days.

Given Today is 20^th January 2017, Sunday
In january, we have 31 days
February - 28 days (Non leap year)
March - 31 days
April - 30 days
⇒ Remaining days = 31 - 20 = 11 in Jan
11 in Jan + 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days
More 20 days to complete 90 days ⇒ upto 20^th April
Therefore, after 90 days from today i.e, 20^th Jan 2017 is 20^th Apr 2017.
Now, the day of the week will be
90/7 ⇒ Remainder '6'
As the day starts with '0' on sunday
6 ⇒ Saturday.
Required day is 20^th April, Saturday.

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24^th Nov, 2007 will be 1 day beyond the day on 24
th Nov, 2006.
But, 24^th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10^th Nov, 2006 is one day before 10^th Nov, 2007 i.e. it is Wednesday.

Given year 2003, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 2003 + 11 = 2014

The year 1985 is an ordinary year. So, it has 1 odd day.
So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.
But, 19th June,1984 is Monday
So, 19th June, 1985 is Tuesday.

The day of the week repeats every 7 days.
Given today is Friday. Again Friday is repeated on the 7^th day, 14^th,... on 7 multiple days.
Hence, Friday is on the 63^rd day, as 63 is multiple of 7.
Now, the required day of the week on the 65^th day is Sunday.

We know that,
1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds.
Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600
1 year = 31536000 seconds.
Hence, 10 years = 31536000 x 10 = 315360000 seconds.

Given year 1939, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 1939 + 11 = 1950

We know that the day repeats every 7 days, 14 days, 21 days, .........
So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.
Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.

Calculating Age has 2 conditions. Let your Birthday is on January 1^st.
1. If the month in which you are born is completed in the present year i.e, your birthday, then
Your Age = Present year - Year you are born
As of now, present year = 2018
i.e, Age = 2019 - 1995 = 24 years.
2. If the month in which you are born is not completed in the present year i.e, your birthday, then
Your Age = Last year - Year you are born
As of now, present year = 2019
i.e, Age = 2018 - 1995 = 23 years.

2^nd Jan 1901 means
(1900 years and 2 day)
Now, 1600 years have 0 odd day
300 years have 1 odd day
2 days has 2 odd day
Total no. of odd days = 0 + 1 + 2 = 3 days
Hence, the day on 2^nd Jan 1901 was Wednesday.

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)
Odd days in 1600 years = 0
Odd days in 300 years = 1
92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days
167 days = (23 weeks + 6 days) =6 odd days.
Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.
Given day is Wednesday.

For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year. So, 1600